Integrand size = 29, antiderivative size = 138 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=-\frac {x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}-\frac {2 b n x^{1-m} (f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x^{-m}}{e}\right )}{d e m^2}+\frac {2 b^2 n^2 x^{1-m} (f x)^{-1+m} \operatorname {PolyLog}\left (2,-\frac {d x^{-m}}{e}\right )}{d e m^3} \]
-x^(1-m)*(f*x)^(-1+m)*(a+b*ln(c*x^n))^2/e/m/(d+e*x^m)-2*b*n*x^(1-m)*(f*x)^ (-1+m)*(a+b*ln(c*x^n))*ln(1+d/e/(x^m))/d/e/m^2+2*b^2*n^2*x^(1-m)*(f*x)^(-1 +m)*polylog(2,-d/e/(x^m))/d/e/m^3
Time = 0.51 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\frac {x^{-m} (f x)^m \left (-\frac {m^2 \left (a+b \log \left (c x^n\right )\right )^2}{d+e x^m}-\frac {2 a b m n \log \left (d-d x^m\right )}{d}+\frac {2 b^2 m n \left (n \log (x)-\log \left (c x^n\right )\right ) \log \left (d-d x^m\right )}{d}+\frac {2 b^2 n^2 \left (\frac {1}{2} m^2 \log ^2(x)+\left (-m \log (x)+\log \left (-\frac {e x^m}{d}\right )\right ) \log \left (d+e x^m\right )+\operatorname {PolyLog}\left (2,1+\frac {e x^m}{d}\right )\right )}{d}\right )}{e f m^3} \]
((f*x)^m*(-((m^2*(a + b*Log[c*x^n])^2)/(d + e*x^m)) - (2*a*b*m*n*Log[d - d *x^m])/d + (2*b^2*m*n*(n*Log[x] - Log[c*x^n])*Log[d - d*x^m])/d + (2*b^2*n ^2*((m^2*Log[x]^2)/2 + (-(m*Log[x]) + Log[-((e*x^m)/d)])*Log[d + e*x^m] + PolyLog[2, 1 + (e*x^m)/d]))/d))/(e*f*m^3*x^m)
Time = 0.55 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2777, 2776, 2779, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx\) |
| \(\Big \downarrow \) 2777 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \int \frac {x^{m-1} \left (a+b \log \left (c x^n\right )\right )^2}{\left (e x^m+d\right )^2}dx\) |
| \(\Big \downarrow \) 2776 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {2 b n \int \frac {a+b \log \left (c x^n\right )}{x \left (e x^m+d\right )}dx}{e m}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}\right )\) |
| \(\Big \downarrow \) 2779 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {2 b n \left (\frac {b n \int \frac {\log \left (\frac {d x^{-m}}{e}+1\right )}{x}dx}{d m}-\frac {\log \left (\frac {d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d m}\right )}{e m}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}\right )\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {2 b n \left (\frac {b n \operatorname {PolyLog}\left (2,-\frac {d x^{-m}}{e}\right )}{d m^2}-\frac {\log \left (\frac {d x^{-m}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d m}\right )}{e m}-\frac {\left (a+b \log \left (c x^n\right )\right )^2}{e m \left (d+e x^m\right )}\right )\) |
x^(1 - m)*(f*x)^(-1 + m)*(-((a + b*Log[c*x^n])^2/(e*m*(d + e*x^m))) + (2*b *n*(-(((a + b*Log[c*x^n])*Log[1 + d/(e*x^m)])/(d*m)) + (b*n*PolyLog[2, -(d /(e*x^m))])/(d*m^2)))/(e*m))
3.4.64.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :> Simp[f^m*(d + e*x^r)^(q + 1)*((a + b*L og[c*x^n])^p/(e*r*(q + 1))), x] - Simp[b*f^m*n*(p/(e*r*(q + 1))) Int[(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d , e, f, m, n, q, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || G tQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_)*(x_))^(m_.)*((d_) + ( e_.)*(x_)^(r_))^(q_.), x_Symbol] :> Simp[(f*x)^m/x^m Int[x^m*(d + e*x^r)^ q*(a + b*Log[c*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] && !(IntegerQ[m] || GtQ[f, 0])
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
\[\int \frac {\left (f x \right )^{m -1} {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{\left (d +e \,x^{m}\right )^{2}}d x\]
Time = 0.30 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.93 \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\frac {{\left (b^{2} e m^{2} n^{2} \log \left (x\right )^{2} + 2 \, {\left (b^{2} e m^{2} n \log \left (c\right ) + a b e m^{2} n\right )} \log \left (x\right )\right )} f^{m - 1} x^{m} - {\left (b^{2} d m^{2} \log \left (c\right )^{2} + 2 \, a b d m^{2} \log \left (c\right ) + a^{2} d m^{2}\right )} f^{m - 1} - 2 \, {\left (b^{2} e f^{m - 1} n^{2} x^{m} + b^{2} d f^{m - 1} n^{2}\right )} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) - 2 \, {\left ({\left (b^{2} e m n \log \left (c\right ) + a b e m n\right )} f^{m - 1} x^{m} + {\left (b^{2} d m n \log \left (c\right ) + a b d m n\right )} f^{m - 1}\right )} \log \left (e x^{m} + d\right ) - 2 \, {\left (b^{2} e f^{m - 1} m n^{2} x^{m} \log \left (x\right ) + b^{2} d f^{m - 1} m n^{2} \log \left (x\right )\right )} \log \left (\frac {e x^{m} + d}{d}\right )}{d e^{2} m^{3} x^{m} + d^{2} e m^{3}} \]
((b^2*e*m^2*n^2*log(x)^2 + 2*(b^2*e*m^2*n*log(c) + a*b*e*m^2*n)*log(x))*f^ (m - 1)*x^m - (b^2*d*m^2*log(c)^2 + 2*a*b*d*m^2*log(c) + a^2*d*m^2)*f^(m - 1) - 2*(b^2*e*f^(m - 1)*n^2*x^m + b^2*d*f^(m - 1)*n^2)*dilog(-(e*x^m + d) /d + 1) - 2*((b^2*e*m*n*log(c) + a*b*e*m*n)*f^(m - 1)*x^m + (b^2*d*m*n*log (c) + a*b*d*m*n)*f^(m - 1))*log(e*x^m + d) - 2*(b^2*e*f^(m - 1)*m*n^2*x^m* log(x) + b^2*d*f^(m - 1)*m*n^2*log(x))*log((e*x^m + d)/d))/(d*e^2*m^3*x^m + d^2*e*m^3)
\[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\int \frac {\left (f x\right )^{m - 1} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x^{m}\right )^{2}}\, dx \]
\[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \left (f x\right )^{m - 1}}{{\left (e x^{m} + d\right )}^{2}} \,d x } \]
2*a*b*f^m*n*(log(x)/(d*e*f*m) - log(e*x^m + d)/(d*e*f*m^2)) - (f^m*log(x^n )^2/(e^2*f*m*x^m + d*e*f*m) - integrate((e*f^m*m*x^m*log(c)^2 + 2*(d*f^m*n + (e*f^m*m*log(c) + e*f^m*n)*x^m)*log(x^n))/(e^3*f*m*x*x^(2*m) + 2*d*e^2* f*m*x*x^m + d^2*e*f*m*x), x))*b^2 - 2*a*b*f^m*log(c*x^n)/(e^2*f*m*x^m + d* e*f*m) - a^2*f^m/(e^2*f*m*x^m + d*e*f*m)
\[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \left (f x\right )^{m - 1}}{{\left (e x^{m} + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(f x)^{-1+m} \left (a+b \log \left (c x^n\right )\right )^2}{\left (d+e x^m\right )^2} \, dx=\int \frac {{\left (f\,x\right )}^{m-1}\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x^m\right )}^2} \,d x \]